A football is kicked at a speed of `20 m//s` a projection angle of `45^(@)`. A receiver on the goal line ` 25` metres away in the direction to the kink runs the same instant to meet the ball. Befrom it hits the ground?

Explanation

Here, u=20 m//s,
`theta =45^(@), d=25 m`.
Herizontal range ,
`R=u^(2)/g sin 2 theta = (20^(2))/(9.8) sin 2 xx 45^(@) = 40.82 m`
Time of flight , `T =(2 u sin theta)/g =(20^(2))/99.8) xx sin 45^(@)`
`=2.886 s `
Since the goal man is already `25 m` away in the direction of the ball so to catch the ball, he is to cover a distane `=40. 82 -25 =15.92 m`, in time `2. 886 s`. Therfore the velocity of goal manto catch
the ball is ` v= (15.82)/(2.886) =5. 481 ms^(-`).